# 6.3. Probability Using Tree Diagrams and Combinations

**Probability Using Tree Diagrams and Combinations**

In this section, we will apply previously learnt counting techniques in calculating probabilities, and use tree diagrams to help us gain a better understanding of what is involved.

We begin with an example.

Example 6.3.1

**Solution**

*E*be the event that the first marble drawn is red, and let

*F*be the event that the second marble drawn is red.

*S*consists of 49 ordered pairs. Of the 49 ordered pairs, there are 3 × 3 = 9 ordered pairs that show red on the first draw and, also, red on the second draw. Therefore:

Example 6.3.2

**Solution**

*S*consists of 42 ordered pairs. Of the 42 ordered pairs, there are 3 × 2 = 6 ordered pairs that show red on the first draw and red on the second draw. Therefore,

*P*(

*E*), and 2/6 represents the probability of drawing a red on the second draw, given that the first draw resulted in a red. We write the latter as

*P*(Red on the second | red on first) or . The “|” represents the word “given.” Therefore:

We now demonstrate the above results with a tree diagram.

Example 6.3.3

**a.**The probability that both marbles are white.

**b.**The probability that the first marble is red and the second white.

**c.**The probability that one marble is red and the other white.

**Solution**

*R*be the event that the marble drawn is red, and let

*W*be the event that the marble drawn is white. We draw the following tree diagram:

Example 6.3.4

**a.**

*P*(Two red and one white)

**b.**

*P*(One of each color)

**c.**

*P*(None blue)

**d.**

*P*(At least one blue)

**Solution**

*R*.

_{1}, R_{2}, R_{3}, W_{1}, W_{2}, B_{1}, B_{2}, B_{3}**a.**

*P*(Two red and one white)

*P*(Two red and one white) = .

**b.**

*P*(One of each color)

*P*(One of each color) = .

**c.**

*P*(None blue)

*P*(None blue) = .

**d.**

*P*(At least one blue)

*P*(At least one blue) =

*P*(one blue, two non-blue) +

*P*(two blue, one non-blue) +

*P*(three blue)

*P*(At least one blue) =

*P*(At least one blue) = 30/56 + 15/56 + 1/56 = 46/56 = 23/28.

*P*(

*E*) = 1 −

*P*(

*E*).

^{c}*E*= At least one blue, then

*E*= None blue. But from part

^{c}**c**of this example, we have (

*E*) = 5/28. Therefore:

^{c}*P*(

*E*) = 1 − 5/28 = 23/28.

Example 6.3.5

**Solution**

*P*(A pair of kings and queens) =

*P*(Two pairs) =

We end the section by solving a problem called the **Birthday Problem**.

Example 6.3.6

**Solution**

*E*represent that at least two people have the same birthday. We first find the probability that no two people have the same birthday. We analyze as follows.

^{25}possible birthdays for 25 people. Therefore, the sample space has 365

^{25}elements. We are interested in the probability that no two people have the same birthday. There are 365 possible choices for the first person and since the second person must have a different birthday, there are 364 choices for the second, 363 for the third, and so on. Therefore:

*P*(No two have the same birthday) =

*P*(at least two people have the same birthday) = 1 −

*P*(No two have the same birthday):

*P*(at least two people have the same birthday) =

# Practice questions

**1. **Two apples are chosen from a basket containing five red and three yellow apples. Draw a tree diagram and find the following probabilities:

**a.** *P *(both red)

**b.** *P *(both yellow)

**2. **Three marbles are drawn from a jar containing five red, four white, and three blue marbles. Find the following probabilities using combinations:

**a.** *P *(all three red)

**b.** *P *(none white)

**3. **A committee of four is selected from a total of 4 occupational and public health students, 5 nursing students, and 6 nutrition students. Find the probabilities for the following events:

**a.** At least three occupational and public health students

**b.** All four students of the same program

**c. **Exactly three students of the same program

Answers

**1. ** **a.**

** b.**

**2. ** **a.**

** b.**

**3. ** **a.**

** b.**

** c.**